Proving there is a primitive root mod $p$ such that $a^{p-1} \not\equiv 1 \pmod{p^2}$ 
Prove there is a primitive root mod $p$ such that $a^{p-1} \not\equiv 1 \pmod{p^2}$.
  ($p$ is an odd prime.)

I don't know how to start this question.
 A: If $(a,p)=1$, then $a^{p-1}\not\equiv 1\pmod {p^2}$ if and only if $a^p\not\equiv a\pmod {p^2}$.
If $a$ is a primite root mod $p$, assume $a^p\equiv a\pmod {p^2}$, or we are done.
Then consider $a'=a+p$. Is $(a')^p \equiv a'\pmod {p^2}$?
Advanced technique/more general result
The group theory/abstract algebra point of view.
If $C_k$ is the cyclic group of order $k$ and $m,n>1$ are relatively prime, then for any exact sequence of abelian groups:
$$0\to C_m\to A \to C_n\to 0$$
we can prove that $A\cong C_{mn}$.
This can be used to prove your theorem by showing an exact sequence:
$$0\to(\mathbb Z_p,+)\to\mathbb Z_{p^2}^\times \to \mathbb Z_{p}^\times\to 0$$
Which yields $m=p$ and $n=p-1$.
A: This is a special case of Hensel lifting. Let $a$ be any number relatively prime to $p$.
For any $t$, we have, by the Binomial Theorem 
$$(a+tp)^{p-1}\equiv a^{p-1}+(p-1)a^{p-2}pt\pmod{p^2}.$$
We have $a^{p-1}\equiv 1\pmod{p}$. Thus as $t$ ranges from $0$ to $p-1$, the number $(a+tp)^{p-1}$ ranges, modulo $p^2$, over all $x\equiv 1\pmod{p}$. In particular, there is exactly one value of $t$ such that $(a+tp)^{p-1}\equiv 1\pmod{p^2}$.
Let $a$ be a primitive root of $p$. Then $a+tp$ is a primitive root of $p$. There is only one $t$ in the interval from $0$ to $p-1$ for which  $(a+tp)^{p-1}   \equiv 1\pmod{p^2}$.
